Let the six numbers a_1, a_2, a_3, a_4, a_5, | vedclass

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Question

$a_1+a_3=10=a_1+d \Rightarrow 5$

$a_1+a_2+a_3+a_4+a_5+a_6=57$

$\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57$

$\Rightarrow a_1+a_6=19$

$\

Vedclass · Accepted Answer

$210$

Vedclass · Answer

$a_1+a_3=10=a_1+d \Rightarrow 5$

$a_1+a_2+a_3+a_4+a_5+a_6=57$

$\Rightarrow \frac{6}{2}\left[a_1+a_6ight]=57$

$\Rightarrow a_1+a_6=19$

$\Rightarrow 2 a_1+5 d=19 	ext { and } a_1+d=5$

$\Rightarrow a_1=2, d=3$

$	ext { Numbers }: 2,5,8,11,14,17$

$	ext { Variance }=\sigma^2=	ext { mean of squares }-	ext { square of mean }$ $=\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}ight)^2 ~\ =\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$

$8 \sigma^2=210$

$x_i$	$0$	$1$	$5$	$6$	$10$	$12$	$17$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to

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